Ok, now, imagine a particle is at a point with coordinates (x, y, z) at some instant t according to the frame S. Since the particle is moving, it will, in general, have different coordinates at an infinitesimally later time t + dt. Due to the finite speed of propagation of the particle, each of the coordinates has to change by an infinitesimal amount. Hence, the coordinates of the particle at a later instant are (x + dx, y + dy, z + dz). By definition, we call the following ratios:
<br />
v_{x} = \frac{dx}{dt}, \; v_{y} = \frac{dy}{dt}, \; v_{x} = \frac{dz}{dt}<br />
the components of the velocity of the particle (relative to the frame S).
Using the Lorentz transformation, we may write for the infinitesimal increments of the coordinates and time in the frame S':
<br />
dx' = (x' + dx') - x' = \gamma \, \left[(x + dx) - \beta \, c \, (t + dt) \right] - \gamma \left(x + \beta \, c \, t \right) = \gamma \, \left(dx - \beta \, c \, dt \right)<br />
<br />
dy' = (y' + dy') - y' = (y + dy) - y = dy<br />
and similarly
<br />
dz' = dz<br />
and, finally:
<br />
dt' = (t' + dt') - t' = \gamma \, \left[(t + dt) - \frac{\beta}{c} \, (x + dx) \right] - \gamma \, \left( t + \frac{\beta}{c} \, x \right) = \gamma \, \left(dt - \frac{\beta}{c} \, dx\right)<br />
I have written everything out, but notice that what we are doing is expressing the differentials of x', y', z', t' as functions of x, y, z, t according to the rules of Multi-Variate Calculus.
Then, according to S', the components of the velocity are:
<br />
v'_{x} = \frac{dx'}{dt'}, \; v'_{y} = \frac{dy'}{dt'}, \; v'_{z} = \frac{dz'}{dt'}<br />
Substituting for all of the above differentials, we get:
<br />
v'_{x} = \frac{dx - \beta \, c \, dt}{dt - \frac{\beta}{c} \, dx}<br />
<br />
v'_{y} = \frac{dy}{\gamma \, \left(dt - \frac{\beta}{c} \, dx \right)}<br />
and, similarly for v'_{z}:
<br />
v'_{z} = \frac{dz}{\gamma \, \left(dt - \frac{\beta}{c} \, dx \right)}<br />
In all of these equations \beta = \frac{u}{c}, where u is the relative velocity of the frame S' with respect to S and \gamma = (1 - \beta^{2})^{-1/2}.
If you divide each term in both the numerator and the denominator by dt and use the definitions for the components of the velocity in the frame S, then you will get:
<br />
v'_{x} = \frac{v_{x} - u}{1 - \frac{v_{x} \, u}{c^{2}}}<br />
<br />
v'_{y} = \frac{v_{y} \sqrt{1 - \frac{u^{2}}{c^{2}}}}{1 - \frac{v_{x} \, u}{c^{2}}}<br />
<br />
v'_{z} = \frac{v_{z} \sqrt{1 - \frac{u^{2}}{c^{2}}}}{1 - \frac{v_{x} \, u}{c^{2}}}<br />
These are the transformation formulas for the components of the velocity. Notice that the formula for the z-component is the same as for the y-component by making the substitution y \rightarrow z everywhere. You can get the inverse transformations (from S' to S) in two ways. Either solve the above equations for the unprimed velocity components or simply use the same formulas with the place of the unprimed and primed components interchanged and u \rightarrow -u. You should get the same result.
Now comes your question:
1. Find the differential dv'_{x} (assuming u is constant);
2. Find the differentials dv'_{y} and then dv'_{z} by analogy;
3. Use the definition of acceleration in S' and S and the above expression for dt' to find the acceleration components transformation formulas.
